1. **Problem 1:** Given $f(x) = x^3 - 5x$, find values of $a$ such that $f'(a) = 13$. 2. **Step 1:** Find the derivative $f'(x)$. $$f'(x) = 3x^2 - 5$$ 3. **Step 2:** Set $f'(a) = 13$ and solve for $a$. $$3a^2 - 5 = 13$$ $$3a^2 = 18$$ $$a^2 = 6$$ $$a = \pm \sqrt{6}$$ 4. **Answer for Problem 1:** $a = \pm \sqrt{6}$. --- 5. **Problem 2:** Given $g(x) = x f(3x + 1)$, $f(4) = 1$, and $f'(4) = 2$, find $g'(1)$. 6. **Step 1:** Use the product rule for derivatives: $$g'(x) = f(3x + 1) + x \cdot f'(3x + 1) \cdot 3$$ 7. **Step 2:** Substitute $x = 1$: $$g'(1) = f(4) + 1 \cdot f'(4) \cdot 3 = 1 + 1 \cdot 2 \cdot 3 = 1 + 6 = 7$$ 8. **Answer for Problem 2:** $g'(1) = 7$. --- 9. **Problem 3:** Given matrices $$A = \begin{pmatrix}4 & 4 & -2 \\ 2 & 6 & 0 \\ 3x & 1 & 3\end{pmatrix}, \quad B = \begin{pmatrix}4 & 2 \\ 3 & 0 \\ 1 & 5\end{pmatrix}$$ and $C = AB$, with $C_{32} = 11$, find $x$. 10. **Step 1:** Calculate $C_{32}$, the element in 3rd row, 2nd column of $C$: $$C_{32} = (3x)(2) + (1)(0) + (3)(5) = 6x + 0 + 15 = 6x + 15$$ 11. **Step 2:** Set equal to 11 and solve for $x$: $$6x + 15 = 11$$ $$6x = -4$$ $$x = -\frac{4}{6} = -\frac{2}{3}$$ 12. **Answer for Problem 3:** $x = -\frac{2}{3}$. --- **Summary of answers:** - Problem 1: $a = \pm \sqrt{6}$ - Problem 2: $g'(1) = 7$ - Problem 3: $x = -\frac{2}{3}$